62nd Ukrainian National Mathematical Olympiad

Let $l$ intersect $AB,AD,BC,CD$ at the points $M,N,K,L$ and let $E=AC\cap QR$. Since $C$ is the orthocenter $AQR$, $AC\perp QR$, and therefore $l$ is the perpendicular bisector of the segment $PE$ (fig. 15). Since $BC$ and $BA$ are the internal and external bisectors of the angle $DBE$, the points $K$ and $M$ are the midpoints of the arcs PE of the circumscribed circle of BPE. Similarly, the points L and N are the midpoints of the arcs PE of the circumscribed circle of DPE. Then $\angle EKL=\angle EBM=\angle ARQ=\angle ECL$, and hence the points C, L, K, E lie on the same circle. Similarly, A, M, N, E lie on the same circle. It remains to see that $\angle LEM=\angle AEM-\angle CEL=\angle ANM-\angle CKL=\angle ARQ-\angle CRQ=90^{\circ }-\angle QAR$. Similarly, $\angle KEN=90^{\circ }-\angle QAR=\angle LEM$, so the circumcircles of the triangles AMN and CKL are in contact.